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Cauchy-Schwartz inequality revisited
Who? Dragoljub J. Keckic
From? University of Belgrade, Faculty of Mathematics
When? May 17th 2018, 15:00
Recall that...
...thisinequality
n∑
k=1
akbk
≤
n∑
k=1
|ak |21/2 n
∑
k=1
|bk |21/2
...and thisinequality
∫
X
f (t)g(t)dμ(t)
≤∫
X
|f (t)|2 dμ(t)1/2 ∫
X
|g(t)|2 dμ(t)1/2
...as well asmany others...
...are specialcases of
| ⟨x , y ⟩ | ≤ ‖x‖‖y‖ - Cauchy-Schwartz inequality in Hilbertspaces.
Recall that...
...thisinequality
n∑
k=1
akbk
≤
n∑
k=1
|ak |21/2 n
∑
k=1
|bk |21/2
...and thisinequality
∫
X
f (t)g(t)dμ(t)
≤∫
X
|f (t)|2 dμ(t)1/2 ∫
X
|g(t)|2 dμ(t)1/2
...as well asmany others...
...are specialcases of
| ⟨x , y ⟩ | ≤ ‖x‖‖y‖ - Cauchy-Schwartz inequality in Hilbertspaces.
Recall that...
...thisinequality
n∑
k=1
akbk
≤
n∑
k=1
|ak |21/2 n
∑
k=1
|bk |21/2
...and thisinequality
∫
X
f (t)g(t)dμ(t)
≤∫
X
|f (t)|2 dμ(t)1/2 ∫
X
|g(t)|2 dμ(t)1/2
...as well asmany others...
...are specialcases of
| ⟨x , y ⟩ | ≤ ‖x‖‖y‖ - Cauchy-Schwartz inequality in Hilbertspaces.
Recall that...
...thisinequality
n∑
k=1
akbk
≤
n∑
k=1
|ak |21/2 n
∑
k=1
|bk |21/2
...and thisinequality
∫
X
f (t)g(t)dμ(t)
≤∫
X
|f (t)|2 dμ(t)1/2 ∫
X
|g(t)|2 dμ(t)1/2
...as well asmany others...
...are specialcases of
| ⟨x , y ⟩ | ≤ ‖x‖‖y‖ - Cauchy-Schwartz inequality in Hilbertspaces.
Similarly
CS inequalityfor Hilbert
modules
| ⟨x , y ⟩ | ≤ ‖x‖ ⟨y , y ⟩1/2 leads to many inequalities, e.g.
1 Jocic, Proc. Amer. Math. Soc. (1998)2 Jocic, J London Math. Soc. (1999)3 Jocic, J Funct. Anal. (2005)4 Jocic, Krtinic, Moslehian, Math. Inequal. Appl. (2013)5 Jocic, Milosevic, Linear Algebra Appl. (2016)6 Milosevic, Adv. Oper. Theory (2016)7 Jocic, Milosevic, Ðuric, Filomat (2017)8 Jocic, Krtinic, Lazarevic, Melentijevic, Milosevic, Complex
Anal. Oper. Theory (2018)
Similarly
CS inequalityfor Hilbert
modules
| ⟨x , y ⟩ | ≤ ‖x‖ ⟨y , y ⟩1/2 leads to many inequalities, e.g.
1 Jocic, Proc. Amer. Math. Soc. (1998)2 Jocic, J London Math. Soc. (1999)3 Jocic, J Funct. Anal. (2005)4 Jocic, Krtinic, Moslehian, Math. Inequal. Appl. (2013)5 Jocic, Milosevic, Linear Algebra Appl. (2016)6 Milosevic, Adv. Oper. Theory (2016)7 Jocic, Milosevic, Ðuric, Filomat (2017)8 Jocic, Krtinic, Lazarevic, Melentijevic, Milosevic, Complex
Anal. Oper. Theory (2018)
Basic definitions
C∗-algebra a Banach algebra with an (antilinear) involution a 7→ a∗
satisfying ‖a∗a‖ = ‖a‖2.
HilbertC∗-module
over A
(W. Paschke, Trans. Amer. Math. Soc. (1973)) a rightmodule over A with an A-valued inner product x, y 7→ ⟨x , y ⟩that satisfies:
1 ⟨x , x⟩ ≥ 0 - the order in A;2 ⟨y , x⟩ = ⟨x , y ⟩∗;3 ⟨x , y1a1 + y2a2⟩ = ⟨x , y1⟩ a1 + ⟨x , y2⟩ a2.
The norm is given by ‖x‖ = ‖ ⟨x , x⟩1/2 ‖.
More details Monographs: 1) E.C. Lance (Cambridge 1995); 2) Manuilov,Troitsky (Moskva 1995 or AMS 2005)
Basic definitions
C∗-algebra a Banach algebra with an (antilinear) involution a 7→ a∗
satisfying ‖a∗a‖ = ‖a‖2.
HilbertC∗-module
over A
(W. Paschke, Trans. Amer. Math. Soc. (1973)) a rightmodule over A with an A-valued inner product x, y 7→ ⟨x , y ⟩that satisfies:
1 ⟨x , x⟩ ≥ 0 - the order in A;2 ⟨y , x⟩ = ⟨x , y ⟩∗;3 ⟨x , y1a1 + y2a2⟩ = ⟨x , y1⟩ a1 + ⟨x , y2⟩ a2.
The norm is given by ‖x‖ = ‖ ⟨x , x⟩1/2 ‖.
More details Monographs: 1) E.C. Lance (Cambridge 1995); 2) Manuilov,Troitsky (Moskva 1995 or AMS 2005)
Basic definitions
C∗-algebra a Banach algebra with an (antilinear) involution a 7→ a∗
satisfying ‖a∗a‖ = ‖a‖2.
HilbertC∗-module
over A
(W. Paschke, Trans. Amer. Math. Soc. (1973)) a rightmodule over A with an A-valued inner product x, y 7→ ⟨x , y ⟩that satisfies:
1 ⟨x , x⟩ ≥ 0 - the order in A;2 ⟨y , x⟩ = ⟨x , y ⟩∗;3 ⟨x , y1a1 + y2a2⟩ = ⟨x , y1⟩ a1 + ⟨x , y2⟩ a2.
The norm is given by ‖x‖ = ‖ ⟨x , x⟩1/2 ‖.
More details Monographs: 1) E.C. Lance (Cambridge 1995); 2) Manuilov,Troitsky (Moskva 1995 or AMS 2005)
Basic definitions
C∗-algebra a Banach algebra with an (antilinear) involution a 7→ a∗
satisfying ‖a∗a‖ = ‖a‖2.
HilbertC∗-module
over A
(W. Paschke, Trans. Amer. Math. Soc. (1973)) a rightmodule over A with an A-valued inner product x, y 7→ ⟨x , y ⟩that satisfies:
1 ⟨x , x⟩ ≥ 0 - the order in A;2 ⟨y , x⟩ = ⟨x , y ⟩∗;3 ⟨x , y1a1 + y2a2⟩ = ⟨x , y1⟩ a1 + ⟨x , y2⟩ a2.
The norm is given by ‖x‖ = ‖ ⟨x , x⟩1/2 ‖.
More details Monographs: 1) E.C. Lance (Cambridge 1995); 2) Manuilov,Troitsky (Moskva 1995 or AMS 2005)
Basic definitions
C∗-algebra a Banach algebra with an (antilinear) involution a 7→ a∗
satisfying ‖a∗a‖ = ‖a‖2.
HilbertC∗-module
over A
(W. Paschke, Trans. Amer. Math. Soc. (1973)) a rightmodule over A with an A-valued inner product x, y 7→ ⟨x , y ⟩that satisfies:
1 ⟨x , x⟩ ≥ 0 - the order in A;2 ⟨y , x⟩ = ⟨x , y ⟩∗;3 ⟨x , y1a1 + y2a2⟩ = ⟨x , y1⟩ a1 + ⟨x , y2⟩ a2.
The norm is given by ‖x‖ = ‖ ⟨x , x⟩1/2 ‖.
More details Monographs: 1) E.C. Lance (Cambridge 1995); 2) Manuilov,Troitsky (Moskva 1995 or AMS 2005)
Cauchy-Schwartz inequality
Proposition We have | ⟨x , y ⟩ |2 ≤ ‖x‖2 ⟨y , y ⟩ and hence| ⟨x , y ⟩ | ≤ ‖x‖ ⟨y , y ⟩1/2. (The first is stronger since t 7→ t1/2 isoperator monotone.)
Proof Expand ⟨xa− y , xa− y ⟩ ≥ 0 to obtaina∗ ⟨x , y ⟩+ ⟨y , x⟩ a ≤ a∗ ⟨x , x⟩ a+ ⟨y , y ⟩ ≤ a∗a+ ⟨y , y ⟩, whenever‖x‖ ≤ 1⇔⟨x , x⟩ ≤ 1. Then, choose a = ⟨x , y ⟩.Given by Paschke (1973). In books at very beginning.
Corollary x , y ∈M - a Hilbert C∗-module over A. Suppose there is aleft action of A on M, i.e. a homomorphism A ,→ Ba(M). Wehave
‖ ⟨x , ay ⟩ ‖ ≤ ‖x‖‖a‖‖y‖. Remember please! (1)
Proof Follows from the previous Proposition and the fact that anyhomomorphism of C∗-algebras is a contraction.
Cauchy-Schwartz inequality
Proposition We have | ⟨x , y ⟩ |2 ≤ ‖x‖2 ⟨y , y ⟩ and hence| ⟨x , y ⟩ | ≤ ‖x‖ ⟨y , y ⟩1/2. (The first is stronger since t 7→ t1/2 isoperator monotone.)
Proof Expand ⟨xa− y , xa− y ⟩ ≥ 0 to obtaina∗ ⟨x , y ⟩+ ⟨y , x⟩ a ≤ a∗ ⟨x , x⟩ a+ ⟨y , y ⟩ ≤ a∗a+ ⟨y , y ⟩, whenever‖x‖ ≤ 1⇔⟨x , x⟩ ≤ 1. Then, choose a = ⟨x , y ⟩.
Given by Paschke (1973). In books at very beginning.
Corollary x , y ∈M - a Hilbert C∗-module over A. Suppose there is aleft action of A on M, i.e. a homomorphism A ,→ Ba(M). Wehave
‖ ⟨x , ay ⟩ ‖ ≤ ‖x‖‖a‖‖y‖. Remember please! (1)
Proof Follows from the previous Proposition and the fact that anyhomomorphism of C∗-algebras is a contraction.
Cauchy-Schwartz inequality
Proposition We have | ⟨x , y ⟩ |2 ≤ ‖x‖2 ⟨y , y ⟩ and hence| ⟨x , y ⟩ | ≤ ‖x‖ ⟨y , y ⟩1/2. (The first is stronger since t 7→ t1/2 isoperator monotone.)
Proof Expand ⟨xa− y , xa− y ⟩ ≥ 0 to obtaina∗ ⟨x , y ⟩+ ⟨y , x⟩ a ≤ a∗ ⟨x , x⟩ a+ ⟨y , y ⟩ ≤ a∗a+ ⟨y , y ⟩, whenever‖x‖ ≤ 1⇔⟨x , x⟩ ≤ 1. Then, choose a = ⟨x , y ⟩.Given by Paschke (1973). In books at very beginning.
Corollary x , y ∈M - a Hilbert C∗-module over A. Suppose there is aleft action of A on M, i.e. a homomorphism A ,→ Ba(M). Wehave
‖ ⟨x , ay ⟩ ‖ ≤ ‖x‖‖a‖‖y‖. Remember please! (1)
Proof Follows from the previous Proposition and the fact that anyhomomorphism of C∗-algebras is a contraction.
Cauchy-Schwartz inequality
Proposition We have | ⟨x , y ⟩ |2 ≤ ‖x‖2 ⟨y , y ⟩ and hence| ⟨x , y ⟩ | ≤ ‖x‖ ⟨y , y ⟩1/2. (The first is stronger since t 7→ t1/2 isoperator monotone.)
Proof Expand ⟨xa− y , xa− y ⟩ ≥ 0 to obtaina∗ ⟨x , y ⟩+ ⟨y , x⟩ a ≤ a∗ ⟨x , x⟩ a+ ⟨y , y ⟩ ≤ a∗a+ ⟨y , y ⟩, whenever‖x‖ ≤ 1⇔⟨x , x⟩ ≤ 1. Then, choose a = ⟨x , y ⟩.Given by Paschke (1973). In books at very beginning.
Corollary x , y ∈M - a Hilbert C∗-module over A. Suppose there is aleft action of A on M, i.e. a homomorphism A ,→ Ba(M). Wehave
‖ ⟨x , ay ⟩ ‖ ≤ ‖x‖‖a‖‖y‖. Remember please! (1)
Proof Follows from the previous Proposition and the fact that anyhomomorphism of C∗-algebras is a contraction.
Cauchy-Schwartz inequality
Proposition We have | ⟨x , y ⟩ |2 ≤ ‖x‖2 ⟨y , y ⟩ and hence| ⟨x , y ⟩ | ≤ ‖x‖ ⟨y , y ⟩1/2. (The first is stronger since t 7→ t1/2 isoperator monotone.)
Proof Expand ⟨xa− y , xa− y ⟩ ≥ 0 to obtaina∗ ⟨x , y ⟩+ ⟨y , x⟩ a ≤ a∗ ⟨x , x⟩ a+ ⟨y , y ⟩ ≤ a∗a+ ⟨y , y ⟩, whenever‖x‖ ≤ 1⇔⟨x , x⟩ ≤ 1. Then, choose a = ⟨x , y ⟩.Given by Paschke (1973). In books at very beginning.
Corollary x , y ∈M - a Hilbert C∗-module over A. Suppose there is aleft action of A on M, i.e. a homomorphism A ,→ Ba(M). Wehave
‖ ⟨x , ay ⟩ ‖ ≤ ‖x‖‖a‖‖y‖. Remember please! (1)
Proof Follows from the previous Proposition and the fact that anyhomomorphism of C∗-algebras is a contraction.
Examples
StandardHilbert module
l2(A) = x =(ξn)+∞n=1 |
∑+∞n=1 ξ
∗nξn converges in the norm of A
... and its dual l2(A)′ = x = (ξn)+∞n=1 |∑+∞
n=1 ξ∗nξn converges weakly
⟨x , y ⟩ =∑+∞
n=1 ξ∗nηn.
L2(Ω;A) If A has a predual then At ∈ L2 iff∫
ΩA∗tAt dμ(t) < +∞ – the
weak integral⟨A,B⟩ =
∫
ΩA∗tBt dμ(t).
Examples
StandardHilbert module
l2(A) = x =(ξn)+∞n=1 |
∑+∞n=1 ξ
∗nξn converges in the norm of A
... and its dual l2(A)′ = x = (ξn)+∞n=1 |∑+∞
n=1 ξ∗nξn converges weakly
⟨x , y ⟩ =∑+∞
n=1 ξ∗nηn.
L2(Ω;A) If A has a predual then At ∈ L2 iff∫
ΩA∗tAt dμ(t) < +∞ – the
weak integral⟨A,B⟩ =
∫
ΩA∗tBt dμ(t).
Examples
StandardHilbert module
l2(A) = x =(ξn)+∞n=1 |
∑+∞n=1 ξ
∗nξn converges in the norm of A
... and its dual l2(A)′ = x = (ξn)+∞n=1 |∑+∞
n=1 ξ∗nξn converges weakly
⟨x , y ⟩ =∑+∞
n=1 ξ∗nηn.
L2(Ω;A) If A has a predual then At ∈ L2 iff∫
ΩA∗tAt dμ(t) < +∞ – the
weak integral⟨A,B⟩ =
∫
ΩA∗tBt dμ(t).
Examples
StandardHilbert module
l2(A) = x =(ξn)+∞n=1 |
∑+∞n=1 ξ
∗nξn converges in the norm of A
... and its dual l2(A)′ = x = (ξn)+∞n=1 |∑+∞
n=1 ξ∗nξn converges weakly
⟨x , y ⟩ =∑+∞
n=1 ξ∗nηn.
L2(Ω;A) If A has a predual then At ∈ L2 iff∫
ΩA∗tAt dμ(t) < +∞ – the
weak integral
⟨A,B⟩ =∫
ΩA∗tBt dμ(t).
Examples
StandardHilbert module
l2(A) = x =(ξn)+∞n=1 |
∑+∞n=1 ξ
∗nξn converges in the norm of A
... and its dual l2(A)′ = x = (ξn)+∞n=1 |∑+∞
n=1 ξ∗nξn converges weakly
⟨x , y ⟩ =∑+∞
n=1 ξ∗nηn.
L2(Ω;A) If A has a predual then At ∈ L2 iff∫
ΩA∗tAt dμ(t) < +∞ – the
weak integral⟨A,B⟩ =
∫
ΩA∗tBt dμ(t).
Main idea
Elementaryoperators
X 7→∑
k AkXBk – (Lummer, Rosenblum 1959) – variousframeworks
Consider X as an operator, and Ak , Bk as variables!Then
∑
k A∗kXBk = ⟨A,XB⟩!
I.p.t.i.transformer
X 7→∫
ΩA∗tXBt dμ(t) = ... again! · · · = ⟨A,XB⟩
Basicinequalities
Inequality (1) becomes
∑
kA∗kXBk
≤
∑
kA∗kAk
1/2
‖X‖
∑
kB∗kBk
1/2
∫
Ω
A∗tXBt
≤
∫
Ω
A∗tAt
1/2
‖X‖
∫
Ω
B∗tBt
1/2
.
Main idea
Elementaryoperators
X 7→∑
k AkXBk – (Lummer, Rosenblum 1959) – variousframeworksConsider X as an operator, and Ak , Bk as variables!Then
∑
k A∗kXBk = ⟨A,XB⟩!
I.p.t.i.transformer
X 7→∫
ΩA∗tXBt dμ(t) = ... again! · · · = ⟨A,XB⟩
Basicinequalities
Inequality (1) becomes
∑
kA∗kXBk
≤
∑
kA∗kAk
1/2
‖X‖
∑
kB∗kBk
1/2
∫
Ω
A∗tXBt
≤
∫
Ω
A∗tAt
1/2
‖X‖
∫
Ω
B∗tBt
1/2
.
Main idea
Elementaryoperators
X 7→∑
k AkXBk – (Lummer, Rosenblum 1959) – variousframeworksConsider X as an operator, and Ak , Bk as variables!Then
∑
k A∗kXBk = ⟨A,XB⟩!
I.p.t.i.transformer
X 7→∫
ΩA∗tXBt dμ(t) = ... again! · · · = ⟨A,XB⟩
Basicinequalities
Inequality (1) becomes
∑
kA∗kXBk
≤
∑
kA∗kAk
1/2
‖X‖
∑
kB∗kBk
1/2
∫
Ω
A∗tXBt
≤
∫
Ω
A∗tAt
1/2
‖X‖
∫
Ω
B∗tBt
1/2
.
Main idea
Elementaryoperators
X 7→∑
k AkXBk – (Lummer, Rosenblum 1959) – variousframeworksConsider X as an operator, and Ak , Bk as variables!Then
∑
k A∗kXBk = ⟨A,XB⟩!
I.p.t.i.transformer
X 7→∫
ΩA∗tXBt dμ(t) = ... again! · · · = ⟨A,XB⟩
Basicinequalities
Inequality (1) becomes
∑
kA∗kXBk
≤
∑
kA∗kAk
1/2
‖X‖
∑
kB∗kBk
1/2
∫
Ω
A∗tXBt
≤
∫
Ω
A∗tAt
1/2
‖X‖
∫
Ω
B∗tBt
1/2
.
Main idea
Elementaryoperators
X 7→∑
k AkXBk – (Lummer, Rosenblum 1959) – variousframeworksConsider X as an operator, and Ak , Bk as variables!Then
∑
k A∗kXBk = ⟨A,XB⟩!
I.p.t.i.transformer
X 7→∫
ΩA∗tXBt dμ(t) = ... again! · · · = ⟨A,XB⟩
Basicinequalities
Inequality (1) becomes
∑
kA∗kXBk
≤
∑
kA∗kAk
1/2
‖X‖
∑
kB∗kBk
1/2
∫
Ω
A∗tXBt
≤
∫
Ω
A∗tAt
1/2
‖X‖
∫
Ω
B∗tBt
1/2
.
Other norms
SemifiniteW∗-algebra
A C∗-algebra with a predual (can be seen as weakly closedsubalgebras of B(H)), and with a semifinite trace τ.
p-norms ‖a‖p = τ(|a|p)1/p. Lp(A;τ) is the completion ofa ∈ A | ‖a‖p < +∞ w.r.t ‖ · ‖p.L1(A;τ)∗ ∼= L∞(A;τ) := A, Lp(A;τ)∗ ∼= Lq(A;τ) - notsurprisingly.
More norms onB(H)
B(H) is a semifinite W∗-algebra with the usual tr.
Ky Fan norms ‖T‖(k) =∑k
j=1 sj (T ) =∑k
j=1 λj (T∗T )1/2.
Unitarilyinvariant norm
Any norm |||·||| that satisfies |||UTV ||| = |||T ||| for all unitaries Uand V .
Ky Fan dom.property
(1951)
|||T ||| ≤ |||S ||| iff for all n ‖T‖(n) ≤ ‖S‖(n).
Other norms
SemifiniteW∗-algebra
A C∗-algebra with a predual (can be seen as weakly closedsubalgebras of B(H)), and with a semifinite trace τ.
p-norms ‖a‖p = τ(|a|p)1/p. Lp(A;τ) is the completion ofa ∈ A | ‖a‖p < +∞ w.r.t ‖ · ‖p.L1(A;τ)∗ ∼= L∞(A;τ) := A, Lp(A;τ)∗ ∼= Lq(A;τ) - notsurprisingly.
More norms onB(H)
B(H) is a semifinite W∗-algebra with the usual tr.
Ky Fan norms ‖T‖(k) =∑k
j=1 sj (T ) =∑k
j=1 λj (T∗T )1/2.
Unitarilyinvariant norm
Any norm |||·||| that satisfies |||UTV ||| = |||T ||| for all unitaries Uand V .
Ky Fan dom.property
(1951)
|||T ||| ≤ |||S ||| iff for all n ‖T‖(n) ≤ ‖S‖(n).
Other norms
SemifiniteW∗-algebra
A C∗-algebra with a predual (can be seen as weakly closedsubalgebras of B(H)), and with a semifinite trace τ.
p-norms ‖a‖p = τ(|a|p)1/p. Lp(A;τ) is the completion ofa ∈ A | ‖a‖p < +∞ w.r.t ‖ · ‖p.L1(A;τ)∗ ∼= L∞(A;τ) := A, Lp(A;τ)∗ ∼= Lq(A;τ) - notsurprisingly.
More norms onB(H)
B(H) is a semifinite W∗-algebra with the usual tr.
Ky Fan norms ‖T‖(k) =∑k
j=1 sj (T ) =∑k
j=1 λj (T∗T )1/2.
Unitarilyinvariant norm
Any norm |||·||| that satisfies |||UTV ||| = |||T ||| for all unitaries Uand V .
Ky Fan dom.property
(1951)
|||T ||| ≤ |||S ||| iff for all n ‖T‖(n) ≤ ‖S‖(n).
Other norms
SemifiniteW∗-algebra
A C∗-algebra with a predual (can be seen as weakly closedsubalgebras of B(H)), and with a semifinite trace τ.
p-norms ‖a‖p = τ(|a|p)1/p. Lp(A;τ) is the completion ofa ∈ A | ‖a‖p < +∞ w.r.t ‖ · ‖p.L1(A;τ)∗ ∼= L∞(A;τ) := A, Lp(A;τ)∗ ∼= Lq(A;τ) - notsurprisingly.
More norms onB(H)
B(H) is a semifinite W∗-algebra with the usual tr.
Ky Fan norms ‖T‖(k) =∑k
j=1 sj (T ) =∑k
j=1 λj (T∗T )1/2.
Unitarilyinvariant norm
Any norm |||·||| that satisfies |||UTV ||| = |||T ||| for all unitaries Uand V .
Ky Fan dom.property
(1951)
|||T ||| ≤ |||S ||| iff for all n ‖T‖(n) ≤ ‖S‖(n).
Other norms
SemifiniteW∗-algebra
A C∗-algebra with a predual (can be seen as weakly closedsubalgebras of B(H)), and with a semifinite trace τ.
p-norms ‖a‖p = τ(|a|p)1/p. Lp(A;τ) is the completion ofa ∈ A | ‖a‖p < +∞ w.r.t ‖ · ‖p.L1(A;τ)∗ ∼= L∞(A;τ) := A, Lp(A;τ)∗ ∼= Lq(A;τ) - notsurprisingly.
More norms onB(H)
B(H) is a semifinite W∗-algebra with the usual tr.
Ky Fan norms ‖T‖(k) =∑k
j=1 sj (T ) =∑k
j=1 λj (T∗T )1/2.
Unitarilyinvariant norm
Any norm |||·||| that satisfies |||UTV ||| = |||T ||| for all unitaries Uand V .
Ky Fan dom.property
(1951)
|||T ||| ≤ |||S ||| iff for all n ‖T‖(n) ≤ ‖S‖(n).
Other norms
SemifiniteW∗-algebra
A C∗-algebra with a predual (can be seen as weakly closedsubalgebras of B(H)), and with a semifinite trace τ.
p-norms ‖a‖p = τ(|a|p)1/p. Lp(A;τ) is the completion ofa ∈ A | ‖a‖p < +∞ w.r.t ‖ · ‖p.L1(A;τ)∗ ∼= L∞(A;τ) := A, Lp(A;τ)∗ ∼= Lq(A;τ) - notsurprisingly.
More norms onB(H)
B(H) is a semifinite W∗-algebra with the usual tr.
Ky Fan norms ‖T‖(k) =∑k
j=1 sj (T ) =∑k
j=1 λj (T∗T )1/2.
Unitarilyinvariant norm
Any norm |||·||| that satisfies |||UTV ||| = |||T ||| for all unitaries Uand V .
Ky Fan dom.property
(1951)
|||T ||| ≤ |||S ||| iff for all n ‖T‖(n) ≤ ‖S‖(n).
Interpolations
Ky Fan norms The dual norm of ‖ · ‖(n) is ‖ · ‖♯(n) =mx‖ · ‖, (1/n)‖ · ‖1.This means ‖T‖(n) = sp‖S‖♯(n)≤1 tr(TS).
Proposition T and S linear, ‖Tx‖ ≤ ‖Sx‖ for all x ∈ C∞, ‖Tx‖1 ≤ ‖Sx‖1 forall x ∈ C1 imply |||T ||| ≤ |||S |||.
Complexinterpolation
betweenLp(A;τ)
S = z ∈ C | 0 < Re z < 1 a strip. If f : S → L1 + L∞ isholomorphic in S, continuous on S and bounded then‖f (it)‖∞ ≤M0, ‖f (1 + it)‖1 ≤M1 implies‖f (1/p)‖p ≤M
1−1/p0 M
1/p1 .
Proof Using Hadammard three line theorem – S.G. and M.G. Krein(1965) for B(H), Pietsch (1971) for semifinite W∗-algebras.The idea can be tracked back to Dixmier (1953).
Corollary The corresponding Riesz-Thorin theorem.
Interpolations
Ky Fan norms The dual norm of ‖ · ‖(n) is ‖ · ‖♯(n) =mx‖ · ‖, (1/n)‖ · ‖1.This means ‖T‖(n) = sp‖S‖♯(n)≤1 tr(TS).
Proposition T and S linear, ‖Tx‖ ≤ ‖Sx‖ for all x ∈ C∞, ‖Tx‖1 ≤ ‖Sx‖1 forall x ∈ C1 imply |||T ||| ≤ |||S |||.
Complexinterpolation
betweenLp(A;τ)
S = z ∈ C | 0 < Re z < 1 a strip. If f : S → L1 + L∞ isholomorphic in S, continuous on S and bounded then‖f (it)‖∞ ≤M0, ‖f (1 + it)‖1 ≤M1 implies‖f (1/p)‖p ≤M
1−1/p0 M
1/p1 .
Proof Using Hadammard three line theorem – S.G. and M.G. Krein(1965) for B(H), Pietsch (1971) for semifinite W∗-algebras.The idea can be tracked back to Dixmier (1953).
Corollary The corresponding Riesz-Thorin theorem.
Interpolations
Ky Fan norms The dual norm of ‖ · ‖(n) is ‖ · ‖♯(n) =mx‖ · ‖, (1/n)‖ · ‖1.This means ‖T‖(n) = sp‖S‖♯(n)≤1 tr(TS).
Proposition T and S linear, ‖Tx‖ ≤ ‖Sx‖ for all x ∈ C∞, ‖Tx‖1 ≤ ‖Sx‖1 forall x ∈ C1 imply |||T ||| ≤ |||S |||.
Complexinterpolation
betweenLp(A;τ)
S = z ∈ C | 0 < Re z < 1 a strip. If f : S → L1 + L∞ isholomorphic in S, continuous on S and bounded then‖f (it)‖∞ ≤M0, ‖f (1 + it)‖1 ≤M1 implies‖f (1/p)‖p ≤M
1−1/p0 M
1/p1 .
Proof Using Hadammard three line theorem – S.G. and M.G. Krein(1965) for B(H), Pietsch (1971) for semifinite W∗-algebras.The idea can be tracked back to Dixmier (1953).
Corollary The corresponding Riesz-Thorin theorem.
Interpolations
Ky Fan norms The dual norm of ‖ · ‖(n) is ‖ · ‖♯(n) =mx‖ · ‖, (1/n)‖ · ‖1.This means ‖T‖(n) = sp‖S‖♯(n)≤1 tr(TS).
Proposition T and S linear, ‖Tx‖ ≤ ‖Sx‖ for all x ∈ C∞, ‖Tx‖1 ≤ ‖Sx‖1 forall x ∈ C1 imply |||T ||| ≤ |||S |||.
Complexinterpolation
betweenLp(A;τ)
S = z ∈ C | 0 < Re z < 1 a strip. If f : S → L1 + L∞ isholomorphic in S, continuous on S and bounded then‖f (it)‖∞ ≤M0, ‖f (1 + it)‖1 ≤M1 implies‖f (1/p)‖p ≤M
1−1/p0 M
1/p1 .
Proof Using Hadammard three line theorem – S.G. and M.G. Krein(1965) for B(H), Pietsch (1971) for semifinite W∗-algebras.The idea can be tracked back to Dixmier (1953).
Corollary The corresponding Riesz-Thorin theorem.
Interpolations
Ky Fan norms The dual norm of ‖ · ‖(n) is ‖ · ‖♯(n) =mx‖ · ‖, (1/n)‖ · ‖1.This means ‖T‖(n) = sp‖S‖♯(n)≤1 tr(TS).
Proposition T and S linear, ‖Tx‖ ≤ ‖Sx‖ for all x ∈ C∞, ‖Tx‖1 ≤ ‖Sx‖1 forall x ∈ C1 imply |||T ||| ≤ |||S |||.
Complexinterpolation
betweenLp(A;τ)
S = z ∈ C | 0 < Re z < 1 a strip. If f : S → L1 + L∞ isholomorphic in S, continuous on S and bounded then‖f (it)‖∞ ≤M0, ‖f (1 + it)‖1 ≤M1 implies‖f (1/p)‖p ≤M
1−1/p0 M
1/p1 .
Proof Using Hadammard three line theorem – S.G. and M.G. Krein(1965) for B(H), Pietsch (1971) for semifinite W∗-algebras.The idea can be tracked back to Dixmier (1953).
Corollary The corresponding Riesz-Thorin theorem.
Modular conjugation
l2(A)′ 3 x = (ξ1, ... ,ξn, ... ) 7→ x = (ξ∗1 , ... ,ξ∗n
, ... ) have thefollowing properties:
1 axb = b∗xa∗;2 τ(⟨y , x⟩) = τ(⟨x , y ⟩).
Definition Any densely defined mapping x 7→ x that satisfies 1 - 2 weshall call modular conjugation.
Definition x is normal if: (i) ⟨x , x⟩ x = x ⟨x , x⟩; (ii) ⟨x , x⟩ = ⟨x , x⟩.
Proposition x, y normal ⇒ ‖⟨x , ay ⟩ ‖ ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖.
Proof Indeed, x ⟨x , x⟩−1/2, y ⟨y , y ⟩−1/2 have norm 1. Hence⟨x , ay ⟩ =
¬
x ⟨x , x⟩−1/2 , ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 y ⟨y , y ⟩−1/2¶
etc.
Oh! Yes I know ⟨x , x⟩ need not be invertible. Then use (⟨x , x⟩+ ϵ)−1/2 andlet ϵ→ 0 before the end of proof.
Modular conjugation
l2(A)′ 3 x = (ξ1, ... ,ξn, ... ) 7→ x = (ξ∗1 , ... ,ξ∗n
, ... ) have thefollowing properties:
1 axb = b∗xa∗;2 τ(⟨y , x⟩) = τ(⟨x , y ⟩).
Definition Any densely defined mapping x 7→ x that satisfies 1 - 2 weshall call modular conjugation.
Definition x is normal if: (i) ⟨x , x⟩ x = x ⟨x , x⟩; (ii) ⟨x , x⟩ = ⟨x , x⟩.
Proposition x, y normal ⇒ ‖⟨x , ay ⟩ ‖ ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖.
Proof Indeed, x ⟨x , x⟩−1/2, y ⟨y , y ⟩−1/2 have norm 1. Hence⟨x , ay ⟩ =
¬
x ⟨x , x⟩−1/2 , ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 y ⟨y , y ⟩−1/2¶
etc.
Oh! Yes I know ⟨x , x⟩ need not be invertible. Then use (⟨x , x⟩+ ϵ)−1/2 andlet ϵ→ 0 before the end of proof.
Modular conjugation
l2(A)′ 3 x = (ξ1, ... ,ξn, ... ) 7→ x = (ξ∗1 , ... ,ξ∗n
, ... ) have thefollowing properties:
1 axb = b∗xa∗;2 τ(⟨y , x⟩) = τ(⟨x , y ⟩).
Definition Any densely defined mapping x 7→ x that satisfies 1 - 2 weshall call modular conjugation.
Definition x is normal if: (i) ⟨x , x⟩ x = x ⟨x , x⟩; (ii) ⟨x , x⟩ = ⟨x , x⟩.
Proposition x, y normal ⇒ ‖⟨x , ay ⟩ ‖ ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖.
Proof Indeed, x ⟨x , x⟩−1/2, y ⟨y , y ⟩−1/2 have norm 1. Hence⟨x , ay ⟩ =
¬
x ⟨x , x⟩−1/2 , ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 y ⟨y , y ⟩−1/2¶
etc.
Oh! Yes I know ⟨x , x⟩ need not be invertible. Then use (⟨x , x⟩+ ϵ)−1/2 andlet ϵ→ 0 before the end of proof.
Modular conjugation
l2(A)′ 3 x = (ξ1, ... ,ξn, ... ) 7→ x = (ξ∗1 , ... ,ξ∗n
, ... ) have thefollowing properties:
1 axb = b∗xa∗;2 τ(⟨y , x⟩) = τ(⟨x , y ⟩).
Definition Any densely defined mapping x 7→ x that satisfies 1 - 2 weshall call modular conjugation.
Definition x is normal if: (i) ⟨x , x⟩ x = x ⟨x , x⟩; (ii) ⟨x , x⟩ = ⟨x , x⟩.
Proposition x, y normal ⇒ ‖⟨x , ay ⟩ ‖ ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖.
Proof Indeed, x ⟨x , x⟩−1/2, y ⟨y , y ⟩−1/2 have norm 1. Hence⟨x , ay ⟩ =
¬
x ⟨x , x⟩−1/2 , ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 y ⟨y , y ⟩−1/2¶
etc.
Oh! Yes I know ⟨x , x⟩ need not be invertible. Then use (⟨x , x⟩+ ϵ)−1/2 andlet ϵ→ 0 before the end of proof.
Modular conjugation
l2(A)′ 3 x = (ξ1, ... ,ξn, ... ) 7→ x = (ξ∗1 , ... ,ξ∗n
, ... ) have thefollowing properties:
1 axb = b∗xa∗;2 τ(⟨y , x⟩) = τ(⟨x , y ⟩).
Definition Any densely defined mapping x 7→ x that satisfies 1 - 2 weshall call modular conjugation.
Definition x is normal if: (i) ⟨x , x⟩ x = x ⟨x , x⟩; (ii) ⟨x , x⟩ = ⟨x , x⟩.
Proposition x, y normal ⇒ ‖⟨x , ay ⟩ ‖ ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖.
Proof Indeed, x ⟨x , x⟩−1/2, y ⟨y , y ⟩−1/2 have norm 1. Hence⟨x , ay ⟩ =
¬
x ⟨x , x⟩−1/2 , ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 y ⟨y , y ⟩−1/2¶
etc.
Oh! Yes I know ⟨x , x⟩ need not be invertible. Then use (⟨x , x⟩+ ϵ)−1/2 andlet ϵ→ 0 before the end of proof.
Modular conjugation
l2(A)′ 3 x = (ξ1, ... ,ξn, ... ) 7→ x = (ξ∗1 , ... ,ξ∗n
, ... ) have thefollowing properties:
1 axb = b∗xa∗;2 τ(⟨y , x⟩) = τ(⟨x , y ⟩).
Definition Any densely defined mapping x 7→ x that satisfies 1 - 2 weshall call modular conjugation.
Definition x is normal if: (i) ⟨x , x⟩ x = x ⟨x , x⟩; (ii) ⟨x , x⟩ = ⟨x , x⟩.
Proposition x, y normal ⇒ ‖⟨x , ay ⟩ ‖ ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖.
Proof Indeed, x ⟨x , x⟩−1/2, y ⟨y , y ⟩−1/2 have norm 1. Hence⟨x , ay ⟩ =
¬
x ⟨x , x⟩−1/2 , ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 y ⟨y , y ⟩−1/2¶
etc.
Oh! Yes I know ⟨x , x⟩ need not be invertible. Then use (⟨x , x⟩+ ϵ)−1/2 andlet ϵ→ 0 before the end of proof.
Inequalities in ‖ · ‖1
Proposition ‖ ⟨x , ay ⟩ ‖1 ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖1.
Proof τ(b ⟨x , ay ⟩) = τ(⟨xb∗, ay ⟩) = τ(⟨ya∗,bx⟩) = τ(a ⟨y ,bx⟩).
Therefore, if ⟨x , x⟩, ⟨y , y ⟩ ≤ 1, then
‖ ⟨x , ay ⟩ ‖1 = sp‖b‖=1
|τ(b ⟨x , ay ⟩)| =
= sp‖b‖=1
|τ(a ⟨y ,bx⟩)| ≤ ‖a‖1‖ ⟨y ,bx⟩ ‖ ≤ ‖a‖1.
In general case x1 = ⟨x , x⟩−1/2 x, y1 = ⟨y , y ⟩−1/2 x, satisfy⟨x1, x1⟩ = 1, ⟨y1, y1⟩ = 1.
Hence ‖ ⟨x , ay ⟩ ‖1 = ‖¬
⟨x , x⟩1/2 x1, a ⟨y , y ⟩1/2 y1¶
‖1 ≤‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖1
Inequalities in ‖ · ‖1
Proposition ‖ ⟨x , ay ⟩ ‖1 ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖1.
Proof τ(b ⟨x , ay ⟩) = τ(⟨xb∗, ay ⟩) = τ(⟨ya∗,bx⟩) = τ(a ⟨y ,bx⟩).
Therefore, if ⟨x , x⟩, ⟨y , y ⟩ ≤ 1, then
‖ ⟨x , ay ⟩ ‖1 = sp‖b‖=1
|τ(b ⟨x , ay ⟩)| =
= sp‖b‖=1
|τ(a ⟨y ,bx⟩)| ≤ ‖a‖1‖ ⟨y ,bx⟩ ‖ ≤ ‖a‖1.
In general case x1 = ⟨x , x⟩−1/2 x, y1 = ⟨y , y ⟩−1/2 x, satisfy⟨x1, x1⟩ = 1, ⟨y1, y1⟩ = 1.
Hence ‖ ⟨x , ay ⟩ ‖1 = ‖¬
⟨x , x⟩1/2 x1, a ⟨y , y ⟩1/2 y1¶
‖1 ≤‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖1
Inequalities in ‖ · ‖1
Proposition ‖ ⟨x , ay ⟩ ‖1 ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖1.
Proof τ(b ⟨x , ay ⟩) = τ(⟨xb∗, ay ⟩) = τ(⟨ya∗,bx⟩) = τ(a ⟨y ,bx⟩).
Therefore, if ⟨x , x⟩, ⟨y , y ⟩ ≤ 1, then
‖ ⟨x , ay ⟩ ‖1 = sp‖b‖=1
|τ(b ⟨x , ay ⟩)| =
= sp‖b‖=1
|τ(a ⟨y ,bx⟩)| ≤ ‖a‖1‖ ⟨y ,bx⟩ ‖ ≤ ‖a‖1.
In general case x1 = ⟨x , x⟩−1/2 x, y1 = ⟨y , y ⟩−1/2 x, satisfy⟨x1, x1⟩ = 1, ⟨y1, y1⟩ = 1.
Hence ‖ ⟨x , ay ⟩ ‖1 = ‖¬
⟨x , x⟩1/2 x1, a ⟨y , y ⟩1/2 y1¶
‖1 ≤‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖1
Inequalities in ‖ · ‖1
Proposition ‖ ⟨x , ay ⟩ ‖1 ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖1.
Proof τ(b ⟨x , ay ⟩) = τ(⟨xb∗, ay ⟩) = τ(⟨ya∗,bx⟩) = τ(a ⟨y ,bx⟩).
Therefore, if ⟨x , x⟩, ⟨y , y ⟩ ≤ 1, then
‖ ⟨x , ay ⟩ ‖1 = sp‖b‖=1
|τ(b ⟨x , ay ⟩)| =
= sp‖b‖=1
|τ(a ⟨y ,bx⟩)| ≤ ‖a‖1‖ ⟨y ,bx⟩ ‖ ≤ ‖a‖1.
In general case x1 = ⟨x , x⟩−1/2 x, y1 = ⟨y , y ⟩−1/2 x, satisfy⟨x1, x1⟩ = 1, ⟨y1, y1⟩ = 1.
Hence ‖ ⟨x , ay ⟩ ‖1 = ‖¬
⟨x , x⟩1/2 x1, a ⟨y , y ⟩1/2 y1¶
‖1 ≤‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖1
Inequalities in ‖ · ‖1
Proposition ‖ ⟨x , ay ⟩ ‖1 ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖1.
Proof τ(b ⟨x , ay ⟩) = τ(⟨xb∗, ay ⟩) = τ(⟨ya∗,bx⟩) = τ(a ⟨y ,bx⟩).
Therefore, if ⟨x , x⟩, ⟨y , y ⟩ ≤ 1, then
‖ ⟨x , ay ⟩ ‖1 = sp‖b‖=1
|τ(b ⟨x , ay ⟩)| =
= sp‖b‖=1
|τ(a ⟨y ,bx⟩)| ≤ ‖a‖1‖ ⟨y ,bx⟩ ‖ ≤ ‖a‖1.
In general case x1 = ⟨x , x⟩−1/2 x, y1 = ⟨y , y ⟩−1/2 x, satisfy⟨x1, x1⟩ = 1, ⟨y1, y1⟩ = 1.
Hence ‖ ⟨x , ay ⟩ ‖1 = ‖¬
⟨x , x⟩1/2 x1, a ⟨y , y ⟩1/2 y1¶
‖1 ≤‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖1
Inequalities in |||·|||
Proposition x and y normal implies
|||⟨x , ay ⟩||| ≤
⟨x , x⟩1/2 a ⟨y , y ⟩1/2
.
Proof Normality ⇒ ⟨x , x⟩ = ⟨x , x⟩ as well as ⟨x , x⟩ x = x ⟨x , x⟩.
Therefore, both ‖ ⟨x , ay ⟩ ‖ ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖ and‖ ⟨x , ay ⟩ ‖1 ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖1. Then interpolate to anyu.i. norm.
Corollaries M = l2(A)′ – the first main result of PAMS 1998.M = L2(Ω;A) – the first main result of JFA 2005.
Inequalities in |||·|||
Proposition x and y normal implies
|||⟨x , ay ⟩||| ≤
⟨x , x⟩1/2 a ⟨y , y ⟩1/2
.
Proof Normality ⇒ ⟨x , x⟩ = ⟨x , x⟩ as well as ⟨x , x⟩ x = x ⟨x , x⟩.
Therefore, both ‖ ⟨x , ay ⟩ ‖ ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖ and‖ ⟨x , ay ⟩ ‖1 ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖1. Then interpolate to anyu.i. norm.
Corollaries M = l2(A)′ – the first main result of PAMS 1998.M = L2(Ω;A) – the first main result of JFA 2005.
Inequalities in |||·|||
Proposition x and y normal implies
|||⟨x , ay ⟩||| ≤
⟨x , x⟩1/2 a ⟨y , y ⟩1/2
.
Proof Normality ⇒ ⟨x , x⟩ = ⟨x , x⟩ as well as ⟨x , x⟩ x = x ⟨x , x⟩.
Therefore, both ‖ ⟨x , ay ⟩ ‖ ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖ and‖ ⟨x , ay ⟩ ‖1 ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖1. Then interpolate to anyu.i. norm.
Corollaries M = l2(A)′ – the first main result of PAMS 1998.M = L2(Ω;A) – the first main result of JFA 2005.
Inequalities in |||·|||
Proposition x and y normal implies
|||⟨x , ay ⟩||| ≤
⟨x , x⟩1/2 a ⟨y , y ⟩1/2
.
Proof Normality ⇒ ⟨x , x⟩ = ⟨x , x⟩ as well as ⟨x , x⟩ x = x ⟨x , x⟩.
Therefore, both ‖ ⟨x , ay ⟩ ‖ ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖ and‖ ⟨x , ay ⟩ ‖1 ≤ ‖ ⟨x , x⟩1/2 a ⟨y , y ⟩1/2 ‖1. Then interpolate to anyu.i. norm.
Corollaries M = l2(A)′ – the first main result of PAMS 1998.M = L2(Ω;A) – the first main result of JFA 2005.
Inequalities in ‖ · ‖p
Idea Since ‖¬
x ⟨x , x⟩−1/2 , ay ⟨y , y ⟩−1/2¶
‖ ≤ ‖a‖ and
‖¬
⟨x , x⟩−1/2 x , a⟨y , y ⟩−1/2 y¶
‖1 ≤ ‖a‖1, consider
z 7→¬
⟨x , x⟩−z/2 x ⟨x , x⟩(z−1)/2 , a ⟨y , y ⟩−z/2 x ⟨y , y ⟩(z−1)/2¶
and apply complex interpolation.
Proposition If 1/ r + 1/q = 2/p then
‖ ⟨x , ay ⟩ ‖p ≤
⟨x , x⟩q−1 x , x1/2q
a
⟨y , y ⟩r−1 y , y1/2r
p.
Corollaries p = r = q, M = l2(A)′, A = B(H) the main result of JLMS1999.M = L2(Ω;A), A = B(H) the second main result of JFA2005.
Inequalities in ‖ · ‖p
Idea Since ‖¬
x ⟨x , x⟩−1/2 , ay ⟨y , y ⟩−1/2¶
‖ ≤ ‖a‖ and
‖¬
⟨x , x⟩−1/2 x , a⟨y , y ⟩−1/2 y¶
‖1 ≤ ‖a‖1, consider
z 7→¬
⟨x , x⟩−z/2 x ⟨x , x⟩(z−1)/2 , a ⟨y , y ⟩−z/2 x ⟨y , y ⟩(z−1)/2¶
and apply complex interpolation.
Proposition If 1/ r + 1/q = 2/p then
‖ ⟨x , ay ⟩ ‖p ≤
⟨x , x⟩q−1 x , x1/2q
a
⟨y , y ⟩r−1 y , y1/2r
p.
Corollaries p = r = q, M = l2(A)′, A = B(H) the main result of JLMS1999.M = L2(Ω;A), A = B(H) the second main result of JFA2005.
Inequalities in ‖ · ‖p
Idea Since ‖¬
x ⟨x , x⟩−1/2 , ay ⟨y , y ⟩−1/2¶
‖ ≤ ‖a‖ and
‖¬
⟨x , x⟩−1/2 x , a⟨y , y ⟩−1/2 y¶
‖1 ≤ ‖a‖1, consider
z 7→¬
⟨x , x⟩−z/2 x ⟨x , x⟩(z−1)/2 , a ⟨y , y ⟩−z/2 x ⟨y , y ⟩(z−1)/2¶
and apply complex interpolation.
Proposition If 1/ r + 1/q = 2/p then
‖ ⟨x , ay ⟩ ‖p ≤
⟨x , x⟩q−1 x , x1/2q
a
⟨y , y ⟩r−1 y , y1/2r
p.
Corollaries p = r = q, M = l2(A)′, A = B(H) the main result of JLMS1999.M = L2(Ω;A), A = B(H) the second main result of JFA2005.
Aczel type inequalities
Recall |1 − ⟨x , y ⟩ | ≥ (1 − ‖x‖2)1/2(1 − ‖y‖2)1/2
Proof |(1 − ⟨x , y ⟩)−1| ≤+∞∑
n=0
| ⟨x , y ⟩ |n ≤+∞∑
n=0
‖x‖n‖y‖n ≤
+∞∑
n=0
‖x‖2n1/2 +∞
∑
n=0
‖y‖2n1/2
= (1 − ‖x‖2)−1/2(1 − ‖y‖2)−1/2.
Proposition x, y normal and ⟨x , x⟩, ⟨y , y ⟩ < 1 implies
(1 − ⟨x , x⟩)1/2a(1 − ⟨y , y ⟩)1/2
≤ |||a− ⟨x , ay ⟩|||
Proof Denote Ta = ⟨x , ay ⟩. Then T 2a = ⟨x , ⟨x , ay ⟩ y ⟩ = ⟨x ⊗ x , a(y ⊗ y)⟩and by induction T na =
x⊗n, ay⊗n
.Put b = (I − T )−1a. Then:
Aczel type inequalities
Recall |1 − ⟨x , y ⟩ | ≥ (1 − ‖x‖2)1/2(1 − ‖y‖2)1/2
Proof |(1 − ⟨x , y ⟩)−1| ≤+∞∑
n=0
| ⟨x , y ⟩ |n ≤+∞∑
n=0
‖x‖n‖y‖n ≤
+∞∑
n=0
‖x‖2n1/2 +∞
∑
n=0
‖y‖2n1/2
= (1 − ‖x‖2)−1/2(1 − ‖y‖2)−1/2.
Proposition x, y normal and ⟨x , x⟩, ⟨y , y ⟩ < 1 implies
(1 − ⟨x , x⟩)1/2a(1 − ⟨y , y ⟩)1/2
≤ |||a− ⟨x , ay ⟩|||
Proof Denote Ta = ⟨x , ay ⟩. Then T 2a = ⟨x , ⟨x , ay ⟩ y ⟩ = ⟨x ⊗ x , a(y ⊗ y)⟩and by induction T na =
x⊗n, ay⊗n
.Put b = (I − T )−1a. Then:
Aczel type inequalities
Recall |1 − ⟨x , y ⟩ | ≥ (1 − ‖x‖2)1/2(1 − ‖y‖2)1/2
Proof |(1 − ⟨x , y ⟩)−1| ≤+∞∑
n=0
| ⟨x , y ⟩ |n ≤+∞∑
n=0
‖x‖n‖y‖n ≤
+∞∑
n=0
‖x‖2n1/2 +∞
∑
n=0
‖y‖2n1/2
= (1 − ‖x‖2)−1/2(1 − ‖y‖2)−1/2.
Proposition x, y normal and ⟨x , x⟩, ⟨y , y ⟩ < 1 implies
(1 − ⟨x , x⟩)1/2a(1 − ⟨y , y ⟩)1/2
≤ |||a− ⟨x , ay ⟩|||
Proof Denote Ta = ⟨x , ay ⟩. Then T 2a = ⟨x , ⟨x , ay ⟩ y ⟩ = ⟨x ⊗ x , a(y ⊗ y)⟩and by induction T na =
x⊗n, ay⊗n
.Put b = (I − T )−1a. Then:
Aczel type inequalities
Recall |1 − ⟨x , y ⟩ | ≥ (1 − ‖x‖2)1/2(1 − ‖y‖2)1/2
Proof |(1 − ⟨x , y ⟩)−1| ≤+∞∑
n=0
| ⟨x , y ⟩ |n ≤+∞∑
n=0
‖x‖n‖y‖n ≤
+∞∑
n=0
‖x‖2n1/2 +∞
∑
n=0
‖y‖2n1/2
= (1 − ‖x‖2)−1/2(1 − ‖y‖2)−1/2.
Proposition x, y normal and ⟨x , x⟩, ⟨y , y ⟩ < 1 implies
(1 − ⟨x , x⟩)1/2a(1 − ⟨y , y ⟩)1/2
≤ |||a− ⟨x , ay ⟩|||
Proof Denote Ta = ⟨x , ay ⟩. Then T 2a = ⟨x , ⟨x , ay ⟩ y ⟩ = ⟨x ⊗ x , a(y ⊗ y)⟩and by induction T na =
x⊗n, ay⊗n
.Put b = (I − T )−1a. Then:
Aczel type inequalities – second frame
Proof –continuation |||b||| =
∑
k≥0T ka
=
∑
k≥0
x⊗k , ay⊗k
=
=
·
k≥0x⊗k , a
∑
k≥0y⊗k
¶
≤
≤
·
k≥0x⊗k ,
∑
k≥0x⊗k
¶12 a¬∑
k≥0y⊗k ,
∑
k≥0y⊗k
¶12
.
However∑
n≥0 x⊗n,
∑
n≥0 x⊗n
=∑
n≥0
x⊗n, x⊗n
=∑
n≥0 ⟨x , x⟩n = (1 − ⟨x , x⟩)−1.
Hence
(I − T )−1a
≤
(1 − ⟨x , x⟩)−1/2a(1 − ⟨y , y ⟩)−1/2
. Fewsubstitution, normality and we are done.
Aczel type inequalities – second frame
Proof –continuation |||b||| =
∑
k≥0T ka
=
∑
k≥0
x⊗k , ay⊗k
=
=
·
k≥0x⊗k , a
∑
k≥0y⊗k
¶
≤
≤
·
k≥0x⊗k ,
∑
k≥0x⊗k
¶12 a¬∑
k≥0y⊗k ,
∑
k≥0y⊗k
¶12
.
However∑
n≥0 x⊗n,
∑
n≥0 x⊗n
=∑
n≥0
x⊗n, x⊗n
=∑
n≥0 ⟨x , x⟩n = (1 − ⟨x , x⟩)−1.
Hence
(I − T )−1a
≤
(1 − ⟨x , x⟩)−1/2a(1 − ⟨y , y ⟩)−1/2
. Fewsubstitution, normality and we are done.
Aczel type inequalities – second frame
Proof –continuation |||b||| =
∑
k≥0T ka
=
∑
k≥0
x⊗k , ay⊗k
=
=
·
k≥0x⊗k , a
∑
k≥0y⊗k
¶
≤
≤
·
k≥0x⊗k ,
∑
k≥0x⊗k
¶12 a¬∑
k≥0y⊗k ,
∑
k≥0y⊗k
¶12
.
However∑
n≥0 x⊗n,
∑
n≥0 x⊗n
=∑
n≥0
x⊗n, x⊗n
=∑
n≥0 ⟨x , x⟩n = (1 − ⟨x , x⟩)−1.
Hence
(I − T )−1a
≤
(1 − ⟨x , x⟩)−1/2a(1 − ⟨y , y ⟩)−1/2
. Fewsubstitution, normality and we are done.
Aczel type inequalities – third frame
Corollaries M = L2(Ω,A) Last main result from JFA2005.M = B(H) × B(H), x = (I ,A), y = (I ,B) last main result fromPAMS1998
Morecorollaries
The preceding proof relies on positive coefficients in Taylorexpansion of t 7→ (1 − t)−1.
Idea Drop normality condition and use complex interpolation(obtained earlier)
Proposition ‖∆1−1/qx
a∆1−1/ry
‖p ≤ ‖∆−1/qx (a− ⟨x , ay ⟩)∆−1/ry ‖p, where
∆z =¬
∑+∞n=0 z
⊗n,∑+∞
n=0 z⊗n¶−1/2
, for z ∈ x , y , x , y, and1/q + 1/ r = 2/p.
Proof Combine previous proof with complex interpolation.
Corollary A = B(H), τ = tr the main result of Filomat 2017. Without ⊗,the formulae seem significantly more robust.
Aczel type inequalities – third frame
Corollaries M = L2(Ω,A) Last main result from JFA2005.M = B(H) × B(H), x = (I ,A), y = (I ,B) last main result fromPAMS1998
Morecorollaries
The preceding proof relies on positive coefficients in Taylorexpansion of t 7→ (1 − t)−1.
Idea Drop normality condition and use complex interpolation(obtained earlier)
Proposition ‖∆1−1/qx
a∆1−1/ry
‖p ≤ ‖∆−1/qx (a− ⟨x , ay ⟩)∆−1/ry ‖p, where
∆z =¬
∑+∞n=0 z
⊗n,∑+∞
n=0 z⊗n¶−1/2
, for z ∈ x , y , x , y, and1/q + 1/ r = 2/p.
Proof Combine previous proof with complex interpolation.
Corollary A = B(H), τ = tr the main result of Filomat 2017. Without ⊗,the formulae seem significantly more robust.
Aczel type inequalities – third frame
Corollaries M = L2(Ω,A) Last main result from JFA2005.M = B(H) × B(H), x = (I ,A), y = (I ,B) last main result fromPAMS1998
Morecorollaries
The preceding proof relies on positive coefficients in Taylorexpansion of t 7→ (1 − t)−1.
Idea Drop normality condition and use complex interpolation(obtained earlier)
Proposition ‖∆1−1/qx
a∆1−1/ry
‖p ≤ ‖∆−1/qx (a− ⟨x , ay ⟩)∆−1/ry ‖p, where
∆z =¬
∑+∞n=0 z
⊗n,∑+∞
n=0 z⊗n¶−1/2
, for z ∈ x , y , x , y, and1/q + 1/ r = 2/p.
Proof Combine previous proof with complex interpolation.
Corollary A = B(H), τ = tr the main result of Filomat 2017. Without ⊗,the formulae seem significantly more robust.
Aczel type inequalities – third frame
Corollaries M = L2(Ω,A) Last main result from JFA2005.M = B(H) × B(H), x = (I ,A), y = (I ,B) last main result fromPAMS1998
Morecorollaries
The preceding proof relies on positive coefficients in Taylorexpansion of t 7→ (1 − t)−1.
Idea Drop normality condition and use complex interpolation(obtained earlier)
Proposition ‖∆1−1/qx
a∆1−1/ry
‖p ≤ ‖∆−1/qx (a− ⟨x , ay ⟩)∆−1/ry ‖p, where
∆z =¬
∑+∞n=0 z
⊗n,∑+∞
n=0 z⊗n¶−1/2
, for z ∈ x , y , x , y, and1/q + 1/ r = 2/p.
Proof Combine previous proof with complex interpolation.
Corollary A = B(H), τ = tr the main result of Filomat 2017. Without ⊗,the formulae seem significantly more robust.
Aczel type inequalities – third frame
Corollaries M = L2(Ω,A) Last main result from JFA2005.M = B(H) × B(H), x = (I ,A), y = (I ,B) last main result fromPAMS1998
Morecorollaries
The preceding proof relies on positive coefficients in Taylorexpansion of t 7→ (1 − t)−1.
Idea Drop normality condition and use complex interpolation(obtained earlier)
Proposition ‖∆1−1/qx
a∆1−1/ry
‖p ≤ ‖∆−1/qx (a− ⟨x , ay ⟩)∆−1/ry ‖p, where
∆z =¬
∑+∞n=0 z
⊗n,∑+∞
n=0 z⊗n¶−1/2
, for z ∈ x , y , x , y, and1/q + 1/ r = 2/p.
Proof Combine previous proof with complex interpolation.
Corollary A = B(H), τ = tr the main result of Filomat 2017. Without ⊗,the formulae seem significantly more robust.
Aczel type inequalities – third frame
Corollaries M = L2(Ω,A) Last main result from JFA2005.M = B(H) × B(H), x = (I ,A), y = (I ,B) last main result fromPAMS1998
Morecorollaries
The preceding proof relies on positive coefficients in Taylorexpansion of t 7→ (1 − t)−1.
Idea Drop normality condition and use complex interpolation(obtained earlier)
Proposition ‖∆1−1/qx
a∆1−1/ry
‖p ≤ ‖∆−1/qx (a− ⟨x , ay ⟩)∆−1/ry ‖p, where
∆z =¬
∑+∞n=0 z
⊗n,∑+∞
n=0 z⊗n¶−1/2
, for z ∈ x , y , x , y, and1/q + 1/ r = 2/p.
Proof Combine previous proof with complex interpolation.
Corollary A = B(H), τ = tr the main result of Filomat 2017. Without ⊗,the formulae seem significantly more robust.
Grüß inequality
Classic form
1
b − a
∫ b
a
fg −1
b − a
∫ b
a
f1
b − a
∫ b
a
g
≤
1
4(mx f −min f )(mx g −min g)
Hilbert spaceform
|⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩| ≤ (⟨x , x⟩ − | ⟨x , e⟩ |2)1/2(⟨y , y ⟩ − | ⟨y , e⟩ |2)1/2,where ‖e‖ = 1
(Observe LHS is a (semi)inner product.)
|⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩| ≤ 14 (M − m)(P − p), if x (y) belongs to the
ball with a diameter [me,Me] ([pe,Pe]), i.e.‖x − e(m + M)/2‖ ≤ (M − m)/2.
(Observe semiinner product is stable under x↔ x − ce.)This is all done by S.S. Dragomir (JMAA 1999)
Grüß inequality
Classic form
1
b − a
∫ b
a
fg −1
b − a
∫ b
a
f1
b − a
∫ b
a
g
≤
1
4(mx f −min f )(mx g −min g)
Hilbert spaceform
|⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩| ≤ (⟨x , x⟩ − | ⟨x , e⟩ |2)1/2(⟨y , y ⟩ − | ⟨y , e⟩ |2)1/2,where ‖e‖ = 1
(Observe LHS is a (semi)inner product.)
|⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩| ≤ 14 (M − m)(P − p), if x (y) belongs to the
ball with a diameter [me,Me] ([pe,Pe]), i.e.‖x − e(m + M)/2‖ ≤ (M − m)/2.
(Observe semiinner product is stable under x↔ x − ce.)This is all done by S.S. Dragomir (JMAA 1999)
Grüß inequality
Classic form
1
b − a
∫ b
a
fg −1
b − a
∫ b
a
f1
b − a
∫ b
a
g
≤
1
4(mx f −min f )(mx g −min g)
Hilbert spaceform
|⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩| ≤ (⟨x , x⟩ − | ⟨x , e⟩ |2)1/2(⟨y , y ⟩ − | ⟨y , e⟩ |2)1/2,where ‖e‖ = 1
(Observe LHS is a (semi)inner product.)
|⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩| ≤ 14 (M − m)(P − p), if x (y) belongs to the
ball with a diameter [me,Me] ([pe,Pe]), i.e.‖x − e(m + M)/2‖ ≤ (M − m)/2.
(Observe semiinner product is stable under x↔ x − ce.)This is all done by S.S. Dragomir (JMAA 1999)
Grüß inequality
Classic form
1
b − a
∫ b
a
fg −1
b − a
∫ b
a
f1
b − a
∫ b
a
g
≤
1
4(mx f −min f )(mx g −min g)
Hilbert spaceform
|⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩| ≤ (⟨x , x⟩ − | ⟨x , e⟩ |2)1/2(⟨y , y ⟩ − | ⟨y , e⟩ |2)1/2,where ‖e‖ = 1
(Observe LHS is a (semi)inner product.)
|⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩| ≤ 14 (M − m)(P − p), if x (y) belongs to the
ball with a diameter [me,Me] ([pe,Pe]), i.e.‖x − e(m + M)/2‖ ≤ (M − m)/2.
(Observe semiinner product is stable under x↔ x − ce.)This is all done by S.S. Dragomir (JMAA 1999)
Grüß inequality
Classic form
1
b − a
∫ b
a
fg −1
b − a
∫ b
a
f1
b − a
∫ b
a
g
≤
1
4(mx f −min f )(mx g −min g)
Hilbert spaceform
|⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩| ≤ (⟨x , x⟩ − | ⟨x , e⟩ |2)1/2(⟨y , y ⟩ − | ⟨y , e⟩ |2)1/2,where ‖e‖ = 1
(Observe LHS is a (semi)inner product.)
|⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩| ≤ 14 (M − m)(P − p), if x (y) belongs to the
ball with a diameter [me,Me] ([pe,Pe]), i.e.‖x − e(m + M)/2‖ ≤ (M − m)/2.
(Observe semiinner product is stable under x↔ x − ce.)This is all done by S.S. Dragomir (JMAA 1999)
Grüß inequality – continuation
Proposition If ⟨e, e⟩ = 1 then Φ(x , y) = ⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩ is a semi-innerproduct.|||⟨x , ay ⟩ − ⟨x , e⟩ ⟨e, ay ⟩||| ≤
(⟨x , x⟩ − | ⟨x , e⟩ |2)1/2a(⟨y , y ⟩ − | ⟨y , e⟩ |2)1/2
, provided x, ynormal w.r.t. Φ.|||⟨x , ay ⟩ − ⟨x , e⟩ ⟨e, ay ⟩||| ≤ 1
4 |||a||| |M − m||P − p|, if x and ybelong to balls with diameters [me,Me], [pe,Pe].
Proof Easy.
Corollaries M = L2(Ω,μ), μ(Ω) = 1, e ≡ 1, then
Φ(x , ay) =∫
Ω
x(t)∗ay(t)dμ(t) −∫
Ω
x(t)∗ dμ(t)∫
Ω
ay(t)dμ(t),
– MIA2013 (17 pages) main result.
Grüß inequality – continuation
Proposition If ⟨e, e⟩ = 1 then Φ(x , y) = ⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩ is a semi-innerproduct.|||⟨x , ay ⟩ − ⟨x , e⟩ ⟨e, ay ⟩||| ≤
(⟨x , x⟩ − | ⟨x , e⟩ |2)1/2a(⟨y , y ⟩ − | ⟨y , e⟩ |2)1/2
, provided x, ynormal w.r.t. Φ.|||⟨x , ay ⟩ − ⟨x , e⟩ ⟨e, ay ⟩||| ≤ 1
4 |||a||| |M − m||P − p|, if x and ybelong to balls with diameters [me,Me], [pe,Pe].
Proof Easy.
Corollaries M = L2(Ω,μ), μ(Ω) = 1, e ≡ 1, then
Φ(x , ay) =∫
Ω
x(t)∗ay(t)dμ(t) −∫
Ω
x(t)∗ dμ(t)∫
Ω
ay(t)dμ(t),
– MIA2013 (17 pages) main result.
Grüß inequality – continuation
Proposition If ⟨e, e⟩ = 1 then Φ(x , y) = ⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩ is a semi-innerproduct.|||⟨x , ay ⟩ − ⟨x , e⟩ ⟨e, ay ⟩||| ≤
(⟨x , x⟩ − | ⟨x , e⟩ |2)1/2a(⟨y , y ⟩ − | ⟨y , e⟩ |2)1/2
, provided x, ynormal w.r.t. Φ.|||⟨x , ay ⟩ − ⟨x , e⟩ ⟨e, ay ⟩||| ≤ 1
4 |||a||| |M − m||P − p|, if x and ybelong to balls with diameters [me,Me], [pe,Pe].
Proof Easy.
Corollaries M = L2(Ω,μ), μ(Ω) = 1, e ≡ 1, then
Φ(x , ay) =∫
Ω
x(t)∗ay(t)dμ(t) −∫
Ω
x(t)∗ dμ(t)∫
Ω
ay(t)dμ(t),
– MIA2013 (17 pages) main result.
Refinements
Remark Till now, we use only ‖ ⟨x , ay ⟩ ‖ ≤ ‖x‖‖a‖‖y‖. However, thereis more precise
| ⟨x , ay ⟩ | ≤ ‖x‖
y , a∗ay1/2
. (2)
Corollaries Apply (2) to M = An to prove some results of LAA 2016.Apply (2) to Φ(x , y) = ⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩ to obtain mainresults of CAOT 2018.
Refinements
Remark Till now, we use only ‖ ⟨x , ay ⟩ ‖ ≤ ‖x‖‖a‖‖y‖. However, thereis more precise
| ⟨x , ay ⟩ | ≤ ‖x‖
y , a∗ay1/2
. (2)
Corollaries Apply (2) to M = An to prove some results of LAA 2016.Apply (2) to Φ(x , y) = ⟨x , y ⟩ − ⟨x , e⟩ ⟨e, y ⟩ to obtain mainresults of CAOT 2018.
Thanks for your attention
Completedetails
arXiv:1801.07953"The applications of Cauchy-Schwartz inequality for Hilbertmodules to elementary operators and i.p.t.i. transformers"submitted 2018. January 24th
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